Answer
$AB=I_{2}$
$BA=I_{2}$
They are inverses of each other.
Work Step by Step
If $A$ is an $n\times n$ matrix, then the inverse of $A$ is the $n\times n$ matrix $A^{-1}$ with the following properties:
$A^{-1}A=I_{n}$ and $AA^{-1}=I_{n}$
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$AB=\left[\begin{array}{ll}
2 & -3\\
4 & -7
\end{array}\right]\left[\begin{array}{ll}
7/2 & -3/2\\
2 & -1
\end{array}\right]$
$=\left[\begin{array}{ll}
2(\frac{7}{2})-3(2) & 2(\frac{-3}{2})-3(-1)\\
4(\frac{7}{2})-7(2) & 4(\frac{-3}{2})-7(-1)
\end{array}\right]$
$=\left[\begin{array}{ll}
7-6 & -3+3\\
14-14 & -6+7
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]=I_{2}$
$BA=\left[\begin{array}{ll}
7/2 & -3/2\\
2 & -1
\end{array}\right]\left[\begin{array}{ll}
2 & -3\\
4 & -7
\end{array}\right]$
$=\left[\begin{array}{ll}
\frac{7}{2}(2)-\frac{3}{2}(4) & \frac{7}{2}(-3)-\frac{3}{2}(-7)\\
2(2)-1(4) & 2(-3)-1(-7)
\end{array}\right]$
$=\left[\begin{array}{ll}
7-6 & -\frac{21}{3}+\frac{21}{3}\\
4-4 & -6+7
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]=I_{2}$
$AB=I_{2}$
$BA=I_{2}$
They are inverses of each other.
